Problem:
In β³ABC with AB=12,BC=13, and AC=15, let M be a point on AC such that the incircles of β³ABM and β³BCM have equal radii. Let p and q be positive relatively prime integers such that CMAMβ=qpβ. Find p+q.
Solution:
Let CMAMβ=k and the common altitude of β³AMB and β³CMB be h. Because the radius of the incircle of triangle equals twice its area divided by its perimeter, the ratio of the areas of two triangles with equal inradii is the same as the ratio of their perimeters. Thus 13+CM+BM12+AM+BMβ= 21βCMβ
h21βAMβ
hβ=k. Replacing AM by kβ
CM and solving for BM yields BM=1βk13kβ12β. The fact that BM>0 implies that 1312β<k. Because CMAMβ=k and AM+CM=15, it follows that CM=k+115β and AM= k+115kβ. Applying the Law of Cosines to triangles ABM and BCM and to angles β BMA=Ξ± and β CMB=ΟβΞ± respectively yields
122=AM2+BM2β2AMβ
BMcosΞ±, and
132=BM2+CM2+2BMβ
CMcosΞ±.
Using AM=kβ
CM, multiplying the second equation by k, and adding the two equations yields
132k+122=BM2(k+1)+AM2+CM2k.
Substituting into the above equation produces
169k+144=(1βk13kβ12β)2(k+1)+(k+115kβ)2+(k+115β)2k
Simplifying this equation yields 4k(69k2β112k+44)=0. Its solutions are k=0, k=32β, and k=2322β. Because k>1312β, only the last solution is valid, and so p+q=45β.
The problems on this page are the property of the MAA's American Mathematics Competitions